Fractions With Nontrivial Anomalous Cancellation

I was working on a sequence Neil Sloane wrote (OEIS A291094). This is a sequence that contains denominators of fractions thus. Let n and d be the numerator and the denominator of a fraction respectively such that I can "cancel" a digit D in both n and in d and have the result = n/d. Usually this doesn't work for obvious reasons. 14/48, cancelling the 4s does not leave us with 7/24 but instead with 1/8, which is why we don't cancel digits when we are trying to simplify fractions! Famously, decimal 16/64, cancelling the 6s, actually does work: it does equal 1/4 like it is "supposed to". This is called "anomalous cancellation."

There is a trivial condition, that is the cancellation of trailing zeros. This is discounted regarding what we're interested in.

This sequence ports nicely into the dozenal condition and it might entertain folks here recreationally. The decimal case 16/64 = 1/4 is famous meme-wise; in dozenal the meme would be 1b;/b6; = 1/6.

In the two-digit case, we are looking for solutions to (xb + y)/(zb + x) = y/z. As MathWorld notes, every proper divisor of the base b corresponds to one solution (and so prime bases have no solutions).

In the dozenal case, omega is prime, and so the divisor solutions (1b/b6 = 1/6, 2b/b8 = 1/4, 3b/b9 = 1/3, 5b/ba = 1/2) are the only solutions. In decimal, omega is composite, so while solutions do come this way (19/95 = 1/5, 49/95 = 1/2) they are not the only ones (16/64 = 1/4, 26/65 = 2/5). There are always an even number of solutions, unless the base is an even square (like 4, 16, or 36).

I have contributed the sequence that notes the smallest denominator d with an anomalous proper cancellation (OEIS A292289). Still have to finish the edit (b-file and chart). I will do the numerator this afternoon and have all cases for bases 2 <= b <= 120, generated overnight. Right now I am working on the number of denominators b < d <= b^2 + b (since primes p have smallest anomalous cancellation proper fraction "11/110" thus (p + 1)/(p^2 + p) = "1/10" = 1/p). From this I can sort how many two base-b digit proper anomalous fractions there are. I have a table similar to the one I am posting here that I will add to A292289 and A292288 once this data is done; it might take a day to generate despite parallelizing the function.

This is not the final document: the final will have 120 terms and the number of d < b^2 (two-digit proper fractions) and d <= b^2 + b.